Combustion analysis

Question

A 9.418 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 15.69 grams of CO₂ and 5.138 grams of H₂O are produced.

In a separate experiment, the molar mass is found to be 132.1 g/mol. Determine the empirical formula and the molecular formula of the organic compound.

Emperical formula:

Molecular formula:

Answer 1

From the combustion data, we can find the moles of C H and O.

15.69 g CO2 x 1 mole CO2/44 g x 1 mol C/mole CO2 = 0.3566 moles C

5.238 g H2O x 1 mol H2O/18 g x 2 mol H/mole H2O = 0.582 moles H

Now, to find moles of O, we will convert moles of C and moles of H to grams. Then subtract that from original mass of sample to get mass of O, and then convert that to moles of O.

g C = 0.3566 moles C x 12 g/mol = 4.279 g C

g H = 0.582 moles H x 1 g/mol = 0.582 g H

g O = 9.418 g - 4.279 g - 0.582 g = 4.557 g O

moles O = 4.557 g O x 1 mol/16 g = 0.2848 moles O

To find empirical formula, we want the lowest whole number multiples of elements, so we divide by the lowest value of moles as follows:

0.2848/0.2848 = 1 mol O

0.3566/0.2848 = 1.25 moles C

0.582/0.2848 = 2.0 moles H

To get whole numbers, we can multiply all by 4 to get

4 moles O, 5 moles C and 8 moles H

Empirical formula = C₅H₈O₄

Molar mass C₅H₈O₄ = 132 g/mol

Since the molar mass of the empirical formula is the same as that of the molecular formula, the molecular formula must be the same as the empirical formula.

Molecular formula = C₅H₈O₄